Voltage DropINPUT DATA:Base MVA=28Fault current at 6.6 KV level=40 kASource Voltage=6.6 kVSource fault level (40x1.732x6.6)=457.248 MVASource P.U. impedance (Base MVA / Fault level MVA)=0.0612Resistance of 6.6 kV (1C x 630 sq.mm)=0.06Resistance of 0.1Km cableR=0.006Reactance of 6.6kV 1Cx 630 Sqmm cable per Km.=0.08Reactance of 0.1Km cableX=0.008Impedance of 6.6KV cable ( Sqr( R2+X2)Zc=0.01P.U. impedance of 6.6 KV cable (Zc*Base MVA / KV2)=0.0064279155Resistance of 6.6 kV 3Cx185 Sqmm cable per Km.=0.211Resistance of 0.1Km cableR=0.0211Reactance of 6.6kV 3Cx185 Sqmm cable per Km.=0.097Reactance of 0.1Km cableX=0.0097Impedance of 6.6KV cable ( Sqr( R2+X2)) ZcZc=0.023223P.U. impedance of 6.6 KV cable (Zc*Base MVA / KV2)=0.0149275482Rated LV System Voltage=415Base impedance=0.0061508929Per Phase LV Voltage -V1=240Distribution transformer capacity mVA=2Actual Transformer impedance of 2000kVA=0.0625Transformer P.U. impedance on 2000kVA=0.03125LV Motor detailsMotor kW rating=180Starting P.F.(cos s)=0.4Starting P.F.(sin s)=0.9162Rated Current in Amp.=309.3185895688Motor Starting current (6 * FLC)=1855.9115374129Starting current to full load current ratio=6 (Assumed)Operating P.F.(cos m)=0.88Operating P.F.(sin m)=0.47Motor connecting cableLength of cable per run(km) (Assumed)=0.15Cable size(sq.mm)=1 X 3C x 400 sq.mmNo.of Runs=1Cable resistance(Rc) (ohm/km)=0.1Cable reactance(Xc) (ohm/km)=0.07Resistance of motor feeder cable (Rc)=0.015Reactance of motor feeder cable (Xc)=0.0105Impedance of 1.1 KV cable ( Sqr( R2+X2)) Zc=0.0181P.U. impedance of 1.1 KV cable (Zc*Base MVA / KV2) Ztl=2.9426622151Total Line Impedance(Ztl)=2.9640176789Source impedance in ohms (P.U.imp*Base imp.) Xg=0.00037643466.6 KV cable impedance in ohms(P.U.imp*Base imp.) Xc1(3C x 185 sq.mm)=0.00009181776.6 KV cable impedance in ohms(P.U.imp*Base imp.) Xc2(1C x 630 sq.mm)=0.0000395Dist. Transformer impedance in ohms (P.U.imp * Base imp.) Xt=0.0001922154Resistance of motor feeder cable (Rc)=0.015Reactance of motor feeder cable (Xc)=0.0105Total source reactance Xs ( Xg +Xc1 + Xc2 + Xt)=0.0006999678Calculation of momentary overloading:( Bus A is considered)When the largest LV motor is to be started(duration of starting is less than 1minute) with all other loads already in service, then the total current(starting current Ist + base load current IB) shall not exceed 1.5times FLC of transformer.Conditions of overloading: (As per Clause 4.1a & 4.1b of IS 6600)a) The overloading is not to exceed 1.5times the rated kVA.b) The hot spot temperature is not to exceed 140c under any circumstances.Transformer KVA=2000Transformer FLC=2782.4925568324Transformer FLC x 1.5 times=4173.7388352486Base load in KVA without largest motor=1407.9Base load current prior to starting of largest motor=1958.7356353822Largest LV Motor FLC=309.3185895688Starting current (Ist)=1855.86Starting kVA of the largest Motor (6 * FLC * kV * 1.732)=1332.61812Total kVA on Transformer During Starting of Largest Motor =2741.9 kVATotal current at the time of start=3814.64 ATotal kVA on Transformer at the time ofStart=2741.9 kVA

Calculation of Voltage at motor terminal VMS:

Nominal supply volatge (V1)2 = (VMS cos + Ist Rc)2 + (VMS sin+ Ist Xc + Ist Xs)2

Voltage at the Motor terminal (VMS)=209.5 V

% voltage drop at motor terminal=(V1-VMS) x 100V1

=12.5626043406

% voltage drop at motor terminal=12.5626043406Voltage Drop at Transformer Terminal:Rated Current of Transformer Secondary:=2782.4925568324Fault Current at Transformer secondary (Ifl / Z%)=44512Fault MVA of Transformer (1.732 * Isc * Vl)=31994.33536=31.99 MVAStarting kVA of the largest Motor (6 * FLC * kV * 1.732) without Base load1333.91 kVAVoltage Drop at Transformer Secondary due to Motor Inrush Current Without base load=0.0416920748=4.16%Total kVA on Transformer During Starting of Largest Motor with Base Load=2741.9 kVAVoltage Drop at Transformer Secondary due to Motor Inrush Current With base load=0.0856995599=8.56%

Cable Size CalcCable Size for 180 kW Motor FeederRated current carrying (continuously) capability:Full load current approximately=309 ACurrent carrying capacity of 3C x 400 sq.mm in Air=530 A

De-rated current carrying capacity of 3C x 400 sq.mmDeraring due to Ambient Temp=0.9Derating due to Grouping=0.66Overall Derating Factor =0.59Current Carrying capcpity after derating (530 * .59)=312.7No of Runs Required =309 / 312=0.99=1 RunTherefore, the cable size selected as 1R 3C x 400 sq.mm /XLPE (Al.) is adequate from continuous current carrying point of viewSteady state voltage dropCable size (selected from above calculation)=1 R # 3C x 400 sq.mmCable length (assumed max.) (L)=.150 kMNo. of runs (n)=1Cable resistance /Km (R) per run=0.1Cable reactance /Km (X) per run=0.07Cos F at rated condition=0.88Sin F = 1-cos2F=0.5Max. Full load current (I)=309Voltage drop V {3xIxLx(Rcos+ Xsin)}/n=3.3 VoltsPercentage voltage drop (V/V) X100%=0.7951807229=0.79%voltage drop during startingCos F at Starting condition=0.4Sin F at Starting condition=0.916Voltage drop V {3xIxLx(Rcos+ Xsin)}/n=50.17Percentage voltage drop (V/V) X100%=(50 / 415 ) * 100=12.09%The Volrage drop of 12.09 % during Starting condition is less than the tolerable limit of 15 %.Hence the Cable size is acceptable

Sheet3